∫ [0-->1] (1-x^2)^(3/2) dx
令x=sinu,则dx=cosudu,x：0-->1,u：0-->π/2,此时所有三角函数为正,(1-x^2)^(3/2)=(cosu)^3
代入原式=∫ [0-->π/2] (cosu)^3*cosu du
=∫ [0-->π/2] (cosu)^4 du
=∫ [0-->π/2] (1/2(1+cos2u))^2 du
=1/4∫ [0-->π/2] (1+cos2u)^2 du
=1/4∫ [0-->π/2] (1+2cos2u+(cos2u)^2) du
=1/4∫ [0-->π/2] (1+2cos2u+1/2(1+cos4u)) du
=1/4∫ [0-->π/2] (3/2+2cos2u+1/2cos4u) du
=1/4 (3/2u+sin2u+1/8sin4u) [0-->π/2]
=3π/16